Amplitude, Period, frequency and phase angle

Click For Summary
SUMMARY

The discussion focuses on the superposition of two vibrations represented by the equations 2cos(ωt) and 3cos(ωt + π/4). The resulting vibration can be expressed in the general form n cos(ωt ± α) by applying trigonometric identities and the formula A cos(x) + B sin(x) = R cos(x - α). The calculations yield R = √(A² + B²) and α = arctan(B/A), leading to a resulting vibration of (4 + 3√2)/2 cos(ωt) - (3√2)/2 sin(ωt).

PREREQUISITES
  • Understanding of trigonometric identities
  • Familiarity with the concept of phase angles in vibrations
  • Knowledge of the superposition principle in wave mechanics
  • Basic skills in solving equations involving sine and cosine functions
NEXT STEPS
  • Study the derivation of the formula A cos(x) + B sin(x) = R cos(x - α)
  • Explore the application of trigonometric identities in wave superposition
  • Learn about the implications of phase angles in mechanical vibrations
  • Investigate the effects of varying amplitude and frequency on resultant vibrations
USEFUL FOR

Students and professionals in physics, mechanical engineering, and vibration analysis who are looking to understand the principles of wave superposition and phase relationships in oscillatory systems.

jenney
Messages
1
Reaction score
0
HELP!

totally lost and confused with this question:
A machine is subject to two vibrations at the same time.
one vibration has the form: 2cosωt and the other vibration has the form: 3 cos(ωt+0.785). (0.785 is actually expressed as pi/4)
determine the resulting vibration and express it in the general form of: n cos(ωt±α)
 
Mathematics news on Phys.org
jenney said:
HELP!

totally lost and confused with this question:
A machine is subject to two vibrations at the same time.
one vibration has the form: 2cosωt and the other vibration has the form: 3 cos(ωt+0.785). (0.785 is actually expressed as pi/4)
determine the resulting vibration and express it in the general form of: n cos(ωt±α)

$2\cos(\omega t) + 3\cos \left(\omega t + \dfrac{\pi}{4} \right)$

$2\cos(\omega t) + 3\left[\cos(\omega t)\cos\left(\dfrac{\pi}{4}\right) - \sin(\omega t)\sin\left(\dfrac{\pi}{4}\right)\right]$

$2\cos(\omega t) + \dfrac{3\sqrt{2}}{2}\left[\cos(\omega t) - \sin(\omega t)\right]$

$\dfrac{4+3\sqrt{2}}{2}\cos(\omega t) - \dfrac{3\sqrt{2}}{2}\sin(\omega t)$note $A\cos{x} + B\sin{x} = R\cos(x - \alpha)$, where ...

$R = \sqrt{A^2+B^2}$ and $\alpha = \arctan\left(\dfrac{B}{A}\right)$

... see what you can do from here. Note that the values for $R$ and $\alpha$ are not "nice".
 

Similar threads

  • · Replies 27 ·
Replies
27
Views
2K
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K